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For auxiliary-field covariant gauge fixing, varying the parameter inserts a BRST-exact operator. Its correlations with separated physical BRST-closed operators vanish by the BRST Ward identity. Normalized physical correlators are therefore independent of the parameter under the invariant-measure and boundary assumptions.

The Polyakov action introduces an independent worldsheet metric $h_{ab}$:Its redundancies are worldsheet diffeomorphisms and Weyl transformations. A Polyakov path integral over all metrics without dividing by this gauge volume would count physically equivalent configurations repeatedly. Fixing conformal gauge therefore needs the Faddeev-Popov determinant, not merely the substitution of a flat metric into the action.

After using the Weyl transformation to remove the trace of a metric variation, the infinitesimal worldsheet diffeomorphism ghost operator isThe Faddeev-Popov determinant of this operator is represented by anticommuting fields: a vector Faddeev-Popov ghost $c^a$ and a symmetric traceless Faddeev-Popov antighost field $b^{ab}$. In Euclidean coordinates their worldsheet ghost action is, up to a consistent overall normalization,The two chiral copies of this bc system have conformal weights $(2,-1)$ and anticommutators $\{b_m,c_n\}={\delta }_{m+n,0}$. They are auxiliary fields for the gauge determinant, not additional spacetime particles. Their use is different from the unwanted timelike negative-norm string states of covariant matter quantization.

The chiral holomorphic stress-energy tensor of the bc system isDouble contractions in its operator product expansion give the central charge of reparameterization ghosts $c_{\mathrm{gh}}=-26$. Equivalently, a fermionic bc system with antighost weight $\lambda$ has $c=1-3(2\lambda -1{)}^2$, and $\lambda =2$ gives $-26$. The free coordinate matter contributes $c_{\mathrm{m}}=D$. Thus the total worldsheet Weyl anomaly vanishes for the flat critical bosonic string theory precisely whenThis statement assumes no additional matter or compensating Liouville field theory.

BRST symmetry expresses the gauge symmetry after gauge fixing by replacing its infinitesimal parameter with the Grassmann variable $c$. In one chiral conformal patch, its graded action has the formthere is an analogous barred copy for a closed string. The Grassmann parity makes $s^{2}X=0$: the term from $sc$ cancels the term from applying $s$ to $\partial X$, while the term with $c^2$ vanishes. The gauge-fixed chiral transformations are on shell when their conservation equations are used. At the quantum level, BRST nilpotence also tests the anomaly.

For example, an open string BRST charge can be written using the matter Virasoro generators asUse the ghost oscillator vacuum $b_{n\ge 0}|0{⟩}_{mathrmgh}=c_{n>0}|0{⟩}_{mathrmgh}=0$ and the corresponding oscillator normal ordering. In this convention the last term is $-c_0$ in the critical theory. With a differently shifted ghost $L_0$, the displayed intercept must be shifted consistently; one must not count the same shift twice. Squaring $Q_B$ shows why both critical conditions are needed. Its anomalous terms, in this convention, have coefficientsThe independent cubic and linear terms vanish at $D=26$, $a=1$. The equivalent chiral BRST current is $j_B=c(T_{mathrmm}+T_{mathrmgh}/2)+(3/2){\partial }^{2}c$, with its contour integral defining the charge.

The graded identity $\{Q_B,b_n\}=L_n^{\mathrm{tot}}$ recovers the gauge constraints. In particular, on $|\psi ⟩\otimes |0{⟩}_{mathrmgh}$, BRST-closed imposes $L_{n>0}^{\mathrm{m}}|\psi ⟩=0$ and $(L_0^{\mathrm{m}}-1)|\psi ⟩=0$. Physical states are classes in BRST cohomology,at the appropriate ghost number: one for the usual open-string vertex and two for the closed-string vertex, with the closed-string zero-mode and closed-string level matching conditions also imposed. Since $Q_B^2=0$, an exact shift preserves closure. BRST Ward identities make exact insertions decouple from physical amplitudes and establish gauge independence. Together with the no-ghost theorem for the critical bosonic string, this explains how the auxiliary ghosts enforce gauge invariance while the physical spectrum retains positive norm.

For adjoint-valued quantities write $(X\times Y{)}_a={\epsilon }_{abc}{X}_b{Y}_c$. Substituting the infinitesimal transformation into the gauge field strength and using the Jacobi identity givesA useful way to organize the calculation is $[D_{\mu },D_{\nu }]X=g{F}_{\mu \nu }\times X$: the gauge field strength transforms covariantly because the gauge covariant derivative does. Thus $\delta (F_a^{\mu \nu }{F}_{\mu \nu a})=2g{F}_a^{\mu \nu }{\epsilon }_{abc}{F}_{\mu \nu b}{\lambda }_c=0$ by antisymmetry. This proves gauge invariance of the Yang-Mills action.

The Yang-Mills equations are $D_{\mu }{F}^{\mu \nu }=0$. They determine physical evolution only modulo gauge redundancy: gauge transformations with arbitrary spacetime-dependent parameters take solutions to equivalent solutions. In Hamiltonian mechanics, $A_0$ has no independent quadratic time-derivative term, and the corresponding equation is a Gauss law constraint in gauge theory. A gauge parameter can vanish with its necessary derivatives on the initial slice yet change the later potential energy, so the potential energy itself is not uniquely fixed by initial physical data.

In the uncorrected path integral, integration along gauge orbits overcounts equivalent fields. At the perturbative level the quadratic operator of the kinetic term has gauge zero modes and no inverse, so it does not supply a quantum field theory propagator. Gauge fixing together with the Faddeev-Popov determinant removes this obstruction in the local perturbative construction.

For the odd BRST transformation, products obey the graded Leibniz rule. The adjoint product with Grassmann variables satisfiesIn particular, $c\times c$ is not zero, and $c\times Dc=Dc\times c$. The supplied transformations givesince the even covariant derivative obeys $D(c\times c)=2(Dc)\times c$. For the ghost,The final equality is the graded Jacobi identity; in components it is the Jacobi identity for ${\epsilon }_{abc}$ contracted with the totally antisymmetric product of three ghosts. Also $s^2\bar{c}=-sb=0$ and $s^{2}b=0$. The square of an odd derivation is an even derivation, because its two mixed product terms cancel. Hence $s^2=0$ on every polynomial in the fields, off shell, without imposing field equations.

Let $\Psi =\int d^{d}x[{\partial }^{\mu }{\bar{c}}_a{A}_{\mu a}+\xi {\bar{c}}_a{b}_a/2]$ be the gauge-fixing fermion. Since $sF=gF\times c$, $s{\mathcal{L}}_{\mathrm{YM}}=0$. The BRST-exact operator in ${\mathcal{L}}_q={\mathcal{L}}_{\mathrm{YM}}-s\psi$ has zero variation by BRST nilpotence, soTo see the resulting kinetic terms and signs, expand the exact term with $s\bar{c}=-b$:After integration by parts,The auxiliary Nakanishi-Lautrup field $b$ can be completed into a square and integrated out, giving the gauge-fixing term $-(\partial \cdot A_a{)}^2/(2\xi )$ for $\xi \ne 0$. At $\xi =0$, retain $b$ as the multiplier enforcing the gauge condition rather than divide by $\xi$. The Grassmann Gaussian integral gives the determinant of the gauge-condition operator $-\partial \cdot D$. Field-independent normalizations cancel in the normalized integral; insertions involving $b$ can be handled before integration or through its source terms.

At $g=0$ the gauge-fixed quadratic operator is invertible with the usual prescription, and the ghosts have a quadratic kinetic term. The remaining terms are cubic gauge interactions of order $g$, quartic gauge interactions of order $g^2$, and a ghost-gauge interaction of order $g$. Expanding them and using Wick contractions gives a perturbative expansion for any defined polynomial insertion $X$ of the fields. The exponent is the gauge-fixed action $S_q$, as printed in the PDF; the converted TeX incorrectly substitutes $S_A$.

At $\xi =1$, the free gauge action, up to a boundary term, is $\frac{1}{2}\int A_{\mu a}\square A_a^{\mu }$. Its Fourier kernel is $-p^2{\eta }_{\mu \nu }{\delta }_{ab}$. With the mostly-plus convention used above, the Feynman-gauge adjoint propagator isThis is consistent with the scalar-line convention in Question 2.

Finally define the odd functional $B=\frac{1}{2}\int d^{d}x{\bar{c}}_a{b}_a$. Then ${\partial }_{\xi }{S}_q=-sB$. Differentiating the normalized expectation, including its denominator, givesfor an insertion with no explicit $\xi$ dependence. A physical gauge-invariant insertion $X[A]$ is BRST-closed, since its infinitesimal gauge variation vanishes also when the parameter is replaced by $c$. The assumed BRST Ward identity gives $⟨sB⟩=0$ and $⟨s(XB)⟩=0$. For an even $X$ with $sX=0$, the latter says $⟨XsB⟩=0$. ThereforeThe same conclusion holds for any BRST-closed insertion. Mere color-singlet invariance of an arbitrary ghost-dependent expression need not imply $sX=0$; the physical-observable qualification is essential. The proof uses the invariant-measure assumptions encoded in the stated Ward identity.

A gauge-invariant observable is BRST-closed: replacing its infinitesimal gauge parameter by the ghost gives $s\mathcal{O}=0$. Change the gauge functional continuously, or interpolate between two admissible choices, by changing the gauge-fixing fermion to ${\Psi }_t$. The action changes by the BRST-exact operator ${\partial }_{t}S=s({\partial }_t{\Psi }_t)$.

For a normalized correlator of $\mathcal{O}={\prod }_i{\mathcal{O}}_i$ with each insertion BRST-closed, differentiation of the functional integral givesThe BRST Ward identity says $⟨sX⟩=0$ for an invariant measure and action, with appropriate boundary conditions. Since $s\mathcal{O}=0$, the graded Leibniz rule makes the first insertion an exact variation of $\mathcal{O}{\partial }_t{\Psi }_t$ up to its harmless parity sign, and both terms vanish. ThusPhysical gauge-invariant correlation functions are independent of this gauge condition, even though individual gauge-field and ghost Feynman diagrams change. This argument assumes an admissible perturbative gauge fixing, a BRST symmetry-preserving regulator/measure and no uncanceled boundary contribution. A global failure of those assumptions is not settled by the formal local calculation.

Let $G_1,\dots ,G_r$ be the separated gauge-invariant insertions. Their BRST symmetry variations vanish: the gauge field strength and its gauge covariant derivatives transform by adjoint commutators, and invariant color contractions remove these commutators. Thus $s{G}_i=0$.

Make the infinitesimal change of integration variables $\Phi \mapsto \Phi +ϵs\Phi$ in the normalized expectation of $\mathcal{O}(x){\prod }_i{G}_i$. The action is invariant up to its boundary term and, by assumption, the measure has no Jacobian anomaly. ThereforeSince $s\mathcal{O}$ is generated by the BRST charge, this is the graded BRST Ward identityHere $[Q,\mathcal{O}\}=Q\mathcal{O}-(-1{)}^{|\mathcal{O}|}\mathcal{O}Q$ is the graded commutator. For even $\mathcal{O}$ it is the ordinary commutator printed in the question; for odd $\mathcal{O}$ it is an anticommutator. Separation of the other insertions from $x$ avoids the additional coincident-point contact terms. An operator need not itself be BRST-closed for this identity to hold.

A BRST-exact insertion has zero correlation with physical, gauge-invariant insertions. This is the decoupling of BRST-exact insertions in physical correlation functions. In the associated BRST cohomology, physical information is represented by closed states or operators modulo exact ones, schematically $\ker Q/\mathrm{im}Q$, using the nilpotence of the BRST charge in the anomaly-free theory. Gauge-fixing fields thereby do not supply additional physical observables.

For example, the gauge parameter dependence is exact:Differentiating a normalized physical correlation function with respect to $\xi$ inserts this expression; the BRST Ward identity makes that derivative zero under the same invariant-measure and boundary assumptions. This illustrates gauge-fixing parameter independence from BRST symmetry.

An infinitesimal change of gauge-fixing fermion, $\Psi \mapsto \Psi +\delta \Psi$, changes the action by the BRST-exact term $Q\delta \Psi$. If $\mathcal{O}$ is gauge invariant, then $Q\mathcal{O}=0$. BRST invariance of the measure gives the BRST Ward identity $⟨QX⟩=0$, so the normalized variation isTherefore correlation functions of gauge-invariant operators do not depend on the choice of gauge, provided there is no BRST anomaly or boundary contribution.